A) 14 unit
B) 12 unit
C) 10 unit
D) 8 unit
Correct Answer: A
Solution :
The equations of given lines are \[\overrightarrow{r}=(5\hat{i}+7\hat{j}+3\hat{k})+\lambda (5\hat{i}-16\hat{j}+7\hat{k})\] and\[\overrightarrow{r}=(9\hat{i}+13\hat{j}+13\hat{k})+15\hat{k})+\mu (3\hat{i}+8\hat{j}-5\hat{k})\] On comparing with\[\overrightarrow{r}=\overrightarrow{{{a}_{1}}}+\lambda \overrightarrow{{{b}_{1}}}\] and\[\overrightarrow{r}=\overrightarrow{{{a}_{2}}}+\mu \,\overrightarrow{{{b}_{2}},}\]we get \[\overrightarrow{{{r}_{1}}}=5\hat{i}+7\hat{j}+3\hat{k},\overrightarrow{{{b}_{1}}}=5\hat{i}-16\hat{j}+7\hat{k}\]and \[\overrightarrow{{{a}_{2}}}=9\hat{i}+13\hat{j}+15\hat{k},\overrightarrow{{{b}_{2}}}=3\hat{i}+8\hat{j}-5\hat{k}\] \[\therefore \]Shortest distance \[=\frac{|(\overrightarrow{{{b}_{1}}}\times \overrightarrow{{{b}_{2}}}).(\overrightarrow{{{a}_{1}}}-\overrightarrow{{{a}_{1}}})|}{|\overrightarrow{{{b}_{1}}}\times \overrightarrow{{{b}_{2}}}|}\] \[=\frac{(24\hat{i}+46\hat{j}+88\hat{k}).(4\hat{i}+6\hat{j}+12\hat{k})}{\sqrt{{{(24)}^{2}}+{{(46)}^{2}}+{{(88)}^{2}}}}\] \[=\frac{1428}{102.16}\] \[=14\]unit
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